# How do you find the intervals of increasing and decreasing given y=x^2/(4x+4)?

Nov 23, 2016

#### Explanation:

The derivative of a function $y$ reveals the increasing/decreasing nature of the function:

• If $y ' > 0$, then $y$ is increasing.
• If $y ' < 0$, then $y$ is decreasing.

So, we need to find the derivative of this function. We will need to use the quotient rule, which states that the derivative of the quotient of the functions $u$ and $v$ is:

$y = \frac{u}{v} \text{ "=>" } y ' = \frac{\left(u ' \cdot v\right) - \left(u \cdot v '\right)}{v} ^ 2$

So, for the given function, we see that:

$y ' = \frac{\left(\frac{d}{\mathrm{dx}} {x}^{2}\right) \left(4 x + 4\right) - \left({x}^{2}\right) \left(\frac{d}{\mathrm{dx}} \left(4 x + 4\right)\right)}{4 x + 4} ^ 2$

Through the power rule, $\frac{d}{\mathrm{dx}} {x}^{2} = 2 x$ and $\frac{d}{\mathrm{dx}} \left(4 x + 4\right) = 4$.

$y ' = \frac{\left(2 x\right) \left(4 x + 4\right) - {x}^{2} \left(4\right)}{4 x + 4} ^ 2 = \frac{4 {x}^{2} + 8 x}{4 x + 4} ^ 2$

Taking ${4}^{2}$ from the denominator and $4 x$ from the numerator:

$y ' = \frac{4 x \left(x + 2\right)}{16 {\left(x + 1\right)}^{2}} = \frac{x \left(x + 2\right)}{4 {\left(x + 1\right)}^{2}}$

So, we want to find when this is positive and when this is negative. Approach this by first finding the spots where the derivative could switch from positive to negative. That is, when the derivative equals $0$ or is undefined.

The derivative equals $0$ when $x \left(x + 2\right) = 0$, or at $x = 0$ and $x = - 2$. The derivative is undefined when $4 {\left(x + 1\right)}^{2} = 0$, or at $x = - 1$.

So, we want to examine the sign of the derivative around the three points $x = - 2 , x = - 1 ,$ and $x = 0$.

When $x < - 2$, we see that the numerator is positive. Note that the denominator will always be positive. Thus the whole derivative is positive for $x < - 2$. This means that $y$ is increasing on $\left(- \infty , - 2\right)$.

For $- 2 < x < - 1$, we see that the numerator is now negative. Thus $y ' < 0$ for this interval and $y$ is decreasing on $\left(- 2 , - 1\right)$.

For $- 1 < x < 0$, the numerator is negative again so $y ' < 0$ on this interval and $y$ is decreasing on $\left(- 1 , 0\right)$.

For $x > 0$, the final interval, the numerator is now positive, so $y ' > 0$ and $y$ is increasing on $\left(0 , \infty\right)$.

$\textcolor{red}{\star}$ We can now conclude that $y$ is decreasing on $\left(- 2 , - 1\right) \cup \left(- 1 , 0\right)$ and is increasing on $\left(- \infty , - 2\right) \cup \left(0 , \infty\right)$.

Note that the decreasing intervals cannot be combined into $\left(- 2 , 0\right)$ because there is an asymptote at $x = - 1$.
graph{x^2/(4x+4) [-10, 10, -5, 5]}